Question

Question #f3e45

Answer 1

Let the radius of the circle be `r`
So area if the circle `pir^2`

Area of the segment `AOB=1/4pir^2`

Area `pir^2` forms `2pi` rad at the center.

So area `1/4pir^2` of segment `AOB` will form `(2pi)/(pir^2)*1/4pir^2=pi/2` rad at the center.

Hence `/_AOB=theta=pi/2`

Now

`theta-sintheta=pi/2-sin(pi/2)=pi/2-1`